Simplify and expand the following expression: $ \dfrac{r - 3}{3r + 9}+\dfrac{r + 3}{4r - 5} $
Answer: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(3r + 9)(4r - 5)$ Multiply the first term by $\dfrac{4r - 5}{4r - 5}$ $ \begin{align*} \dfrac{r - 3}{3r + 9} \times \dfrac{4r - 5}{4r - 5} & = \dfrac{(r - 3)(4r - 5)}{(3r + 9)(4r - 5)} \\ & = \dfrac{4r^2 - 17r + 15}{(3r + 9)(4r - 5)}\end{align*} $ Multiply the second term by $\dfrac{3r + 9}{3r + 9}$ $ \begin{align*} \dfrac{r + 3}{4r - 5} \times \dfrac{3r + 9}{3r + 9} & = \dfrac{(r + 3)(3r + 9)}{(4r - 5)(3r + 9)} \\ & = \dfrac{3r^2 + 18r + 27}{(4r - 5)(3r + 9)}\end{align*} $ Now we have: $ = \dfrac{4r^2 - 17r + 15}{(3r + 9)(4r - 5)} + \dfrac{3r^2 + 18r + 27}{(4r - 5)(3r + 9)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{4r^2 - 17r + 15 + 3r^2 + 18r + 27}{(3r + 9)(4r - 5)} $ $ = \dfrac{7r^2 + r + 42}{(3r + 9)(4r - 5)}$ Expand the denominator: $ = \dfrac{7r^2 + r + 42}{12r^2 + 21r - 45}$